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Calculate Money in Leetcode Bank

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Topics:

ArraysGreedy Algorithms

Hercy wants to save money for his first car. He puts money in the Leetcode bank every day.

He starts by putting in $1 on Monday, the first day. Every day from Tuesday to Sunday, he will put in $1 more than the day before. On every subsequent Monday, he will put in $1 more than the previous Monday.

Given n, return the total amount of money he will have in the Leetcode bank at the end of the n^th day.

Example 1:

Input: n = 4
Output: 10
Explanation: After the 4^th day, the total is 1 + 2 + 3 + 4 = 10.

Example 2:

Input: n = 10
Output: 37
Explanation: After the 10^th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2^nd Monday, Hercy only puts in $2.

Example 3:

Input: n = 20
Output: 96
Explanation: After the 20^th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.

Constraints:

1 <= n <= 1000

Try coding it in LeetCode

Calculate Money in Leetcode Bank

Easy

Asked by:

43 views

Topics:

ArraysGreedy Algorithms

Hercy wants to save money for his first car. He puts money in the Leetcode bank every day.

Given n, return the total amount of money he will have in the Leetcode bank at the end of the n^th day.

Example 1:

Input: n = 4
Output: 10
Explanation: After the 4^th day, the total is 1 + 2 + 3 + 4 = 10.

Example 2:

Input: n = 10
Output: 37
Explanation: After the 10^th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2^nd Monday, Hercy only puts in $2.

Example 3:

Input: n = 20
Output: 96
Explanation: After the 20^th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.

Constraints:

1 <= n <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of `n`? Is it within the range of a standard integer, or should I be concerned about potential overflow issues?
Is `n` guaranteed to be a non-negative integer? Should I handle the case where `n` is zero or negative?
Can you provide an example of the expected output for a moderate value of `n`, such as `n = 10` or `n = 14`, to ensure I fully understand the depositing pattern?
Is there a specific data type I should use for the return value, given the potential for the total amount to grow relatively large?
If 'n' is large, is there a preference between a more space-optimized solution versus one that favors more readable code, assuming similar time complexities?

Brute Force Solution

Approach

We need to calculate the total money accumulated in a Leetcode bank over a certain number of days, with deposits increasing each week. The brute force approach simulates the deposits day by day, week by week, following the problem's rules to calculate the money deposited each day and the overall total.

Here's how the algorithm would work step-by-step:

Start with the first day and determine how much money is deposited on that day, which depends on the week number and day of the week.
Add this deposit amount to a running total of all the money in the bank.
Move to the next day and repeat the process of figuring out the deposit amount for that day and adding it to the running total.
Keep doing this for every single day up to the specified number of days.
The final running total will be the total amount of money in the Leetcode bank.

Code Implementation

def calculate_total_money(
    number_of_days: int) -> int:

    total_money = 0
    week_number = 0
    day_of_week = 0

    for day in range(1, number_of_days + 1):
        # Calculate the deposit for the current day
        daily_deposit =
            week_number + day_of_week + 1

        # Add the deposit to the total money
        total_money += daily_deposit

        # Move to the next day of the week
        day_of_week += 1

        # If it's the end of the week, reset
        # the day and increment week
        if day_of_week == 7:
            day_of_week = 0
            week_number += 1

    return total_money

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each day from 1 to n, where n is the number of days. For each day, it performs a constant amount of work to determine the deposit amount based on the week and day of the week, and then adds it to a running total. Since the number of operations grows linearly with the input size n (the number of days), the time complexity is O(n).

Space Complexity

O(1) – The provided algorithm calculates the total money by iterating through each day and adding the corresponding deposit to a running total. It only uses a few constant space variables, such as the running total itself and loop counters to track the week and day. No additional data structures like arrays or hash maps are created. Therefore, the space complexity remains constant regardless of the input 'n' (number of days).

Optimal Solution

Approach

The most efficient way to calculate the total money is to recognize the repeating pattern of how much money is added each week. We can calculate the sum for full weeks, and then add the remaining days from the last week.

Here's how the algorithm would work step-by-step:

Figure out how many full weeks we have.
Calculate the total money earned from all the full weeks. Notice that each week's sum is an arithmetic sequence, and the sums of these sequences themselves form an arithmetic sequence.
Determine how many days are left over after the full weeks.
Calculate the money earned from the remaining days. This is another simple arithmetic sequence.
Add the money earned from the full weeks and the money earned from the remaining days to get the total amount of money.

Code Implementation

def calculate_money(days):
    weeks_completed = days // 7

    # Calculating total for full weeks
    money_from_full_weeks = weeks_completed * (2 * 28 + (weeks_completed - 1) * 7) // 2

    remaining_days = days % 7

    # Calculate total for remaining days
    money_from_remaining_days = (weeks_completed + 1 + weeks_completed + remaining_days) * remaining_days // 2

    # Sum the weekly and remaining days
    total_money = money_from_full_weeks + money_from_remaining_days

    return total_money

Big(O) Analysis

Time Complexity

O(1) – The number of weeks and remaining days are calculated using division and modulo operations on the input n, taking constant time. The calculations for full weeks and remaining days involve arithmetic formulas, which also take constant time regardless of the input size n. Therefore, the entire algorithm executes in a fixed amount of time irrespective of n. This means the time complexity is O(1).

Space Complexity

O(1) – The algorithm primarily uses a few variables to store the number of full weeks, the sum of full weeks, the number of remaining days, and the sum of the remaining days. The number of such variables is constant and does not depend on the input 'n' (the total number of days). Therefore, the space used is constant regardless of the input size, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
n is 0	Return 0 immediately as there are no days to deposit.
n is a small positive number (e.g., 1, 2, 7)	Calculate the sum directly by iterating up to n, handling the first week.
n is a large positive number approaching the maximum integer limit	Use long to prevent integer overflow when calculating the total sum.
n is a multiple of 7	The algorithm should correctly calculate full weeks worth of deposits.
n is slightly larger than a multiple of 7 (e.g., n = 8, 9)	The algorithm should correctly handle the partial week by adding the appropriate values.
Integer overflow potential in the sum calculation	Use a data type with a larger range, such as 'long', to prevent integer overflow during accumulation.
Negative input for n	Throw an IllegalArgumentException or return an error code since the number of days must be non-negative.
n is equal to the maximum integer value	Ensure the calculation of number of weeks and remaining days don't cause an integer overflow.

n is 0

How to Handle:

Return 0 immediately as there are no days to deposit.

n is a small positive number (e.g., 1, 2, 7)

How to Handle:

Calculate the sum directly by iterating up to n, handling the first week.

n is a large positive number approaching the maximum integer limit

How to Handle:

Use long to prevent integer overflow when calculating the total sum.

n is a multiple of 7

How to Handle:

The algorithm should correctly calculate full weeks worth of deposits.

n is slightly larger than a multiple of 7 (e.g., n = 8, 9)

How to Handle:

The algorithm should correctly handle the partial week by adding the appropriate values.

Integer overflow potential in the sum calculation

How to Handle:

Use a data type with a larger range, such as 'long', to prevent integer overflow during accumulation.

Negative input for n

How to Handle:

Throw an IllegalArgumentException or return an error code since the number of days must be non-negative.

n is equal to the maximum integer value

How to Handle:

Ensure the calculation of number of weeks and remaining days don't cause an integer overflow.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of `n`? Is it within the range of a standard integer, or should I be concerned about potential overflow issues?
Is `n` guaranteed to be a non-negative integer? Should I handle the case where `n` is zero or negative?
Can you provide an example of the expected output for a moderate value of `n`, such as `n = 10` or `n = 14`, to ensure I fully understand the depositing pattern?
Is there a specific data type I should use for the return value, given the potential for the total amount to grow relatively large?
If 'n' is large, is there a preference between a more space-optimized solution versus one that favors more readable code, assuming similar time complexities?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the first day and determine how much money is deposited on that day, which depends on the week number and day of the week.
Add this deposit amount to a running total of all the money in the bank.
Move to the next day and repeat the process of figuring out the deposit amount for that day and adding it to the running total.
Keep doing this for every single day up to the specified number of days.
The final running total will be the total amount of money in the Leetcode bank.

Code Implementation

def calculate_total_money(
    number_of_days: int) -> int:

    total_money = 0
    week_number = 0
    day_of_week = 0

    for day in range(1, number_of_days + 1):
        # Calculate the deposit for the current day
        daily_deposit =
            week_number + day_of_week + 1

        # Add the deposit to the total money
        total_money += daily_deposit

        # Move to the next day of the week
        day_of_week += 1

        # If it's the end of the week, reset
        # the day and increment week
        if day_of_week == 7:
            day_of_week = 0
            week_number += 1

    return total_money

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Figure out how many full weeks we have.
Calculate the total money earned from all the full weeks. Notice that each week's sum is an arithmetic sequence, and the sums of these sequences themselves form an arithmetic sequence.
Determine how many days are left over after the full weeks.
Calculate the money earned from the remaining days. This is another simple arithmetic sequence.
Add the money earned from the full weeks and the money earned from the remaining days to get the total amount of money.

Code Implementation

def calculate_money(days):
    weeks_completed = days // 7

    # Calculating total for full weeks
    money_from_full_weeks = weeks_completed * (2 * 28 + (weeks_completed - 1) * 7) // 2

    remaining_days = days % 7

    # Calculate total for remaining days
    money_from_remaining_days = (weeks_completed + 1 + weeks_completed + remaining_days) * remaining_days // 2

    # Sum the weekly and remaining days
    total_money = money_from_full_weeks + money_from_remaining_days

    return total_money

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n is 0	Return 0 immediately as there are no days to deposit.
n is a small positive number (e.g., 1, 2, 7)	Calculate the sum directly by iterating up to n, handling the first week.
n is a large positive number approaching the maximum integer limit	Use long to prevent integer overflow when calculating the total sum.
n is a multiple of 7	The algorithm should correctly calculate full weeks worth of deposits.
n is slightly larger than a multiple of 7 (e.g., n = 8, 9)	The algorithm should correctly handle the partial week by adding the appropriate values.
Integer overflow potential in the sum calculation	Use a data type with a larger range, such as 'long', to prevent integer overflow during accumulation.
Negative input for n	Throw an IllegalArgumentException or return an error code since the number of days must be non-negative.
n is equal to the maximum integer value	Ensure the calculation of number of weeks and remaining days don't cause an integer overflow.

n is 0

How to Handle:

Return 0 immediately as there are no days to deposit.

n is a small positive number (e.g., 1, 2, 7)

How to Handle:

Calculate the sum directly by iterating up to n, handling the first week.

n is a large positive number approaching the maximum integer limit

How to Handle:

Use long to prevent integer overflow when calculating the total sum.

n is a multiple of 7

How to Handle:

The algorithm should correctly calculate full weeks worth of deposits.

n is slightly larger than a multiple of 7 (e.g., n = 8, 9)

How to Handle:

The algorithm should correctly handle the partial week by adding the appropriate values.

Integer overflow potential in the sum calculation

How to Handle:

Use a data type with a larger range, such as 'long', to prevent integer overflow during accumulation.

Negative input for n

How to Handle:

Throw an IllegalArgumentException or return an error code since the number of days must be non-negative.

n is equal to the maximum integer value

How to Handle:

Ensure the calculation of number of weeks and remaining days don't cause an integer overflow.